Vtyjkyuk

Alright, a homework problem.

I'm stuck at this question,

Eliminate $x$ and $y$ from the given equations to get a single
equation in terms of $a$ , $b$ and $c$

$$beginalign
x^2 y &= a \
x(x+y) &= b \
2x+y &=c
endalign$$

Let me tell you what I tried, I tried to get $y$ from one equation and substitute in the other two. Turns out that I'm not able to fully get rid of both $x$ and $y$. Help please.

I'm confused why you say in comments that "still $y$ does not get eliminated" -- at a point in the conversation where $y$ should have disappeared long ago.

You can solve your third equation $2x+y=c$ to get $y=c-2x$. When you plug that into the two other equations you get
$$ x^2(c-2x)=a \
x(x+c-2x)= b $$
No matter what you do subsequently, there won't be any $y$s left to deal with!

What I would do at this point is put off dividing for as long as possible, so rewrite the second equation to
$$ x^2 = cx - b $$
This form lets you reduce any polynomial in $x$ to a first-degree polynomial, by repeatedly using it to eliminate the highest-degree term. We will use it to simplify the third-degree first equation. First substitute the leading $x^2$ to get
$$ (cx-b)(c-2x) = a $$
Multiply out:
$$ c^2x + 2bx - 2cx^2 - bc = a $$
and then insert $x^2=cx-b$ once again:
$$ c^2x + 2bx - 2c(cx-b) - bc = a $$
This is finally a linear equation in $x$. You can solve it without running into any square roots, and insert this value for $x$ into $x^2=cx-b$. What is left then is a rational equation in only $a$, $b$, and $c$, as desired.

Unless the coefficient of $x$ in the linear equation was $0$ (which is a case you'll need to handle separately), each step along the way was reversible, so you will know that if you have $a,b,c$ that satisfy the final equation, they will also have a solution for $x$ and $y$.

Combine the first and second equations to eliminate $y$:

$x(x+fracax^2)=b$

which tidies to

$x^2+fracax=b$ ....... (1)

combine the second and third equations to eliminate $y$:

$x(x+c-2x)=b$

which tidies to

$x^2-cx+b=0$

solving this quadratic gives

$x=fracc pm sqrtc^2-4b2$

which you can substitute into equation (1):

$left(fracc pm sqrtc^2-4b2right)^2+fracaleft(fracc pm sqrtc^2-4b2right)=b$

You can notice that the product resp. the sum of $x$ and $x+y$ are $c$ resp. $b$, you solve that and get a simple system in $x$ and $x+y$ and you substitute in the first equation. Pretty same ideas.

I will work with 2nd and 3rd equations to solve for $x$ and $y$.

From (3) we get,
$$y=c-2x$$
Substituting the value in (2) we get,
$$x(x+(c-2x))=b$$
$$implies x^2-cx+b=0$$
$$x=dfrac12(c+sqrtc^2-4b) text or, dfrac12(c+sqrtc^2-4b)$$
If $x=dfrac12(c+sqrtc^2-4b)$,then,
$$y=-sqrtc^2-4b$$

Now,we plug the values of x and y in equation in (1),
We get,

$$-dfrac14(c^2+2csqrtc^2-4b+c^2-4b)sqrtc^2-4b=a$$
And this is the required equation.

[If you use another value of $x$ then,we will get another equation like it.]